Question: Consider the parametric curve: $\begin{aligned} x&=\sin^2(t) \\\\ y&=-9t \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=0$ to $t=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{0}^{4} \sqrt{4\sin^2(t)+81}\,dt$ (Choice B) B $\int_{0}^{4} \sqrt{4\cos^2(t)+81}\,dt$ (Choice C) C $\int_{0}^{4} \sqrt{\sin^4(t)+81t^2}\,dt$ (Choice D) D $\int_{0}^{4} \sqrt{4\sin^2(t)\cos^2(t)+81}\,dt$
Explanation: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\sin^2(t)\right] \\\\ &=2\sin(t)\cos(t) \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[-9t\right] \\\\ &=-9 \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{0}^{4} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{0}^{4} \sqrt{\left(2\sin(t)\cos(t)\right)^2+\left(-9\right)^2}\,dt \\\\ &=\int_{0}^{4} \sqrt{4\sin^2(t)\cos^2(t)+81}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=0$ to $t=4$ : $\int_{0}^{4} \sqrt{4\sin^2(t)\cos^2(t)+81}\,dt$